+
+ return 0;
+}
+
+namespace {
+ /// \brief Predicate tests whether a ValueEntry's op is in a map.
+ struct IsValueInMap {
+ const DenseMap<Value *, unsigned> ⤅
+
+ IsValueInMap(const DenseMap<Value *, unsigned> &Map) : Map(Map) {}
+
+ bool operator()(const ValueEntry &Entry) {
+ return Map.find(Entry.Op) != Map.end();
+ }
+ };
+}
+
+/// \brief Build up a vector of value/power pairs factoring a product.
+///
+/// Given a series of multiplication operands, build a vector of factors and
+/// the powers each is raised to when forming the final product. Sort them in
+/// the order of descending power.
+///
+/// (x*x) -> [(x, 2)]
+/// ((x*x)*x) -> [(x, 3)]
+/// ((((x*y)*x)*y)*x) -> [(x, 3), (y, 2)]
+///
+/// \returns Whether any factors have a power greater than one.
+bool Reassociate::collectMultiplyFactors(SmallVectorImpl<ValueEntry> &Ops,
+ SmallVectorImpl<Factor> &Factors) {
+ // FIXME: Have Ops be (ValueEntry, Multiplicity) pairs, simplifying this.
+ // Compute the sum of powers of simplifiable factors.
+ unsigned FactorPowerSum = 0;
+ for (unsigned Idx = 1, Size = Ops.size(); Idx < Size; ++Idx) {
+ Value *Op = Ops[Idx-1].Op;
+
+ // Count the number of occurrences of this value.
+ unsigned Count = 1;
+ for (; Idx < Size && Ops[Idx].Op == Op; ++Idx)
+ ++Count;
+ // Track for simplification all factors which occur 2 or more times.
+ if (Count > 1)
+ FactorPowerSum += Count;
+ }
+
+ // We can only simplify factors if the sum of the powers of our simplifiable
+ // factors is 4 or higher. When that is the case, we will *always* have
+ // a simplification. This is an important invariant to prevent cyclicly
+ // trying to simplify already minimal formations.
+ if (FactorPowerSum < 4)
+ return false;
+
+ // Now gather the simplifiable factors, removing them from Ops.
+ FactorPowerSum = 0;
+ for (unsigned Idx = 1; Idx < Ops.size(); ++Idx) {
+ Value *Op = Ops[Idx-1].Op;
+
+ // Count the number of occurrences of this value.
+ unsigned Count = 1;
+ for (; Idx < Ops.size() && Ops[Idx].Op == Op; ++Idx)
+ ++Count;
+ if (Count == 1)
+ continue;
+ // Move an even number of occurrences to Factors.
+ Count &= ~1U;
+ Idx -= Count;
+ FactorPowerSum += Count;
+ Factors.push_back(Factor(Op, Count));
+ Ops.erase(Ops.begin()+Idx, Ops.begin()+Idx+Count);
+ }
+
+ // None of the adjustments above should have reduced the sum of factor powers
+ // below our mininum of '4'.
+ assert(FactorPowerSum >= 4);
+
+ std::sort(Factors.begin(), Factors.end(), Factor::PowerDescendingSorter());
+ return true;
+}
+
+/// \brief Build a tree of multiplies, computing the product of Ops.
+static Value *buildMultiplyTree(IRBuilder<> &Builder,
+ SmallVectorImpl<Value*> &Ops) {
+ if (Ops.size() == 1)
+ return Ops.back();
+
+ Value *LHS = Ops.pop_back_val();
+ do {
+ LHS = Builder.CreateMul(LHS, Ops.pop_back_val());
+ } while (!Ops.empty());
+
+ return LHS;
+}
+
+/// \brief Build a minimal multiplication DAG for (a^x)*(b^y)*(c^z)*...
+///
+/// Given a vector of values raised to various powers, where no two values are
+/// equal and the powers are sorted in decreasing order, compute the minimal
+/// DAG of multiplies to compute the final product, and return that product
+/// value.
+Value *Reassociate::buildMinimalMultiplyDAG(IRBuilder<> &Builder,
+ SmallVectorImpl<Factor> &Factors) {
+ assert(Factors[0].Power);
+ SmallVector<Value *, 4> OuterProduct;
+ for (unsigned LastIdx = 0, Idx = 1, Size = Factors.size();
+ Idx < Size && Factors[Idx].Power > 0; ++Idx) {
+ if (Factors[Idx].Power != Factors[LastIdx].Power) {
+ LastIdx = Idx;
+ continue;
+ }
+
+ // We want to multiply across all the factors with the same power so that
+ // we can raise them to that power as a single entity. Build a mini tree
+ // for that.
+ SmallVector<Value *, 4> InnerProduct;
+ InnerProduct.push_back(Factors[LastIdx].Base);
+ do {
+ InnerProduct.push_back(Factors[Idx].Base);
+ ++Idx;
+ } while (Idx < Size && Factors[Idx].Power == Factors[LastIdx].Power);
+
+ // Reset the base value of the first factor to the new expression tree.
+ // We'll remove all the factors with the same power in a second pass.
+ Value *M = Factors[LastIdx].Base = buildMultiplyTree(Builder, InnerProduct);
+ if (Instruction *MI = dyn_cast<Instruction>(M))
+ RedoInsts.insert(MI);
+
+ LastIdx = Idx;
+ }
+ // Unique factors with equal powers -- we've folded them into the first one's
+ // base.
+ Factors.erase(std::unique(Factors.begin(), Factors.end(),
+ Factor::PowerEqual()),
+ Factors.end());
+
+ // Iteratively collect the base of each factor with an add power into the
+ // outer product, and halve each power in preparation for squaring the
+ // expression.
+ for (unsigned Idx = 0, Size = Factors.size(); Idx != Size; ++Idx) {
+ if (Factors[Idx].Power & 1)
+ OuterProduct.push_back(Factors[Idx].Base);
+ Factors[Idx].Power >>= 1;
+ }
+ if (Factors[0].Power) {
+ Value *SquareRoot = buildMinimalMultiplyDAG(Builder, Factors);
+ OuterProduct.push_back(SquareRoot);
+ OuterProduct.push_back(SquareRoot);
+ }
+ if (OuterProduct.size() == 1)
+ return OuterProduct.front();
+
+ Value *V = buildMultiplyTree(Builder, OuterProduct);
+ return V;
+}
+
+Value *Reassociate::OptimizeMul(BinaryOperator *I,
+ SmallVectorImpl<ValueEntry> &Ops) {
+ // We can only optimize the multiplies when there is a chain of more than
+ // three, such that a balanced tree might require fewer total multiplies.
+ if (Ops.size() < 4)
+ return 0;
+
+ // Try to turn linear trees of multiplies without other uses of the
+ // intermediate stages into minimal multiply DAGs with perfect sub-expression
+ // re-use.
+ SmallVector<Factor, 4> Factors;
+ if (!collectMultiplyFactors(Ops, Factors))
+ return 0; // All distinct factors, so nothing left for us to do.
+
+ IRBuilder<> Builder(I);
+ Value *V = buildMinimalMultiplyDAG(Builder, Factors);
+ if (Ops.empty())
+ return V;
+
+ ValueEntry NewEntry = ValueEntry(getRank(V), V);
+ Ops.insert(std::lower_bound(Ops.begin(), Ops.end(), NewEntry), NewEntry);